1
1 Non-Linear Mathematical Steady State Model of the EPN
For EPN (electric power network) there are usually given the active and
reactive powers in the load nodes.
A steady state in the power system is
uniquely
determined
, if in all nodes
these values are known :
- absolute voltage value
U
,
- voltage angle
,
- active power
P
,
- reactive power
Q
.
Usually two values in the nodes are given and the others can be
determined by solution of steady state calculation.
The nodes of the EPN are divided according to values, which are given for
individual node, as follows:
2
1.1 Node Classification in the EPN
1. Type (
U
,
) – the balance node
This node is usually marked in the equivalent network as the first one. By solution
of steady state we receive the active
P
and reactive power
Q
.
2. Type (
P
,
Q
) – the supply, or load nodes with the given active and reactive power.
The power of supply node and load node differs in a sign. The solution of steady
state are the voltages
U
and their angles
.
3. Type (
P
,
U
) – so called regulation or compensation nodes.
The absolute voltage values and the active power values are given in these nodes.
The solution of steady state is the reactive power
Q
required for keeping the
desired voltage of a node and its angle
.
P
,
U
nodes are used to represent:
- generation nodes with voltage control (by the reactive power),
- nodes with synchronous compensator
- it is synchronous machine without any active
power generation or load (except for losses) used for voltage control (by reactive power)
3
In practical, EPN (with total number of the nodes
n
) is usually given by:
• one balance node (
U
,
),
•
k
nodes (
P
,
Q
),
•
n-k-
1
nodes (
U
,
P
).
given parameters
result of the steady
state calculation
P6
Q6
P1
Q1
P4 Q4
P2
Q2
U1,
1
P5
Q5
U2,
2
U5,
5
U6,
6
P3 Q3
U3,
3
U4,
4
4
1.2 Formulation of the Equations for Steady State Calculation of EPN
With application of the node voltages method
for the given network we can write:
4
3
2
1
44
43
42
41
34
33
32
31
24
23
22
21
14
13
12
11
4
3
2
1
.
U
U
U
U
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
I
I
I
I
By multiplication of this equation we obtain:
4
44
3
43
2
42
1
41
4
4
34
3
33
2
32
1
31
3
4
24
3
23
2
22
1
21
2
4
14
3
13
2
12
1
11
1
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
U
Y
U
Y
U
Y
U
Y
I
U
Y
U
Y
U
Y
U
Y
I
U
Y
U
Y
U
Y
U
Y
I
U
Y
U
Y
U
Y
U
Y
I
unknown
variables
If we can determine unknown variables , it is possible to eliminate the first
equation. We obtain 3 equations with 3 unknown variables.
4
3
2
,
,
U
U
U
known voltage
of the balance
node
5
4
44
3
43
2
42
1
41
4
4
34
3
33
2
32
1
31
3
4
24
3
23
2
22
1
21
2
.
.
.
.
.
.
.
.
.
.
.
.
U
Y
U
Y
U
Y
U
Y
I
U
Y
U
Y
U
Y
U
Y
I
U
Y
U
Y
U
Y
U
Y
I
These equations can be written in form:
4
,
3
,
2
.
4
1
i
U
Y
I
j
j
ij
i
Generally, for the network with
n
nodes we can write:
n
i
U
Y
I
n
j
j
ij
i
,
,
3
,
2
.
1
n
-1 equations with
n
-1 unknown
variables
n
U
U
...,
,
2
In case, that in the
i –
th node is given the active and reactive power, equation for this
power in one phase
*
of the network is:
i
i
i
i
i
I
U
Q
P
S
*
.
j
i
I
*
where is the complex conjugate of the current
i
I
i
i
i
i
i
i
U
Q
P
U
S
I
*
*
*
j
n
i
U
Y
U
Q
P
n
j
j
ij
i
i
i
,
,
3
,
2
.
j
1
*
*
In the next calculation, we suppose that the system (sources, transmission elements and loads) is symmetric, therefore the
problem can be solved as the one-phase network.
6
n
j
j
ij
i
i
i
U
Y
U
Q
P
1
*
.
j
This model is called
Non-Linear mathematical steady state model of EPN
(the powers are given in the nodes).
This expression presents system of
n
-1 non-linear equations with
n
-1 unknown variables.
n
j
j
ij
i
i
i
U
Y
U
Q
P
1
*
.
.
j
In practise, the non-linear equation system is most often solved by: Gauss-Seidl and
Newton iterative method.
n
i
,
,
3
,
2
n
i
,
,
3
,
2